All categories

2 years ago

Is 3/7 bigger than 1/4

Mathematics

1 answer:

dmitriy555 [2]2 years ago

8 0

Yes! it is bigger than

You might be interested in

Please help I need to do this ASAP!!

labwork [276]

Yah I can’t see it either

8 0

2 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

Find the equation of the lien shown.

Mnenie [13.5K]

Answer:

y = $\frac{1}{2}$ x

Step-by-step explanation:

the equation of a line passing through the origin is

y = mx ( m is the slope )

calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, 0) and (x₂, y₂ ) = (6, 3) ← 2 points on the line

m = $\frac{3-0}{6-0}$ = $\frac{3}{6}$ = $\frac{1}{2}$

y = $\frac{1}{2}$ x ← equation of line

7 0

1 year ago

It is given that y varies inversely as x^2. If x is decreased by 50%, find the percentage change in y.

vova2212 [387]

Answer:

300%

Step-by-step explanation:

Given

$y\ \alpha\ \frac{1}{x^2}$

Required

Find the percentage change in y when x decreased by 50%

First, convert to equation

$y\ = \ \frac{k}{x^2}$

Where k is the constant of proportionality

When x decreased by 50%

$Y = \frac{k}{(x - 0.5x)^2}$

$Y = \frac{k}{(0.5x)^2}$

$Y = \frac{k}{0.25x^2}$

Expand

$Y = \frac{1}{0.25} * \frac{k}{x^2}$

Substitute $\frac{k}{x^2}$ for y

$Y = \frac{1}{0.25} * y$

$Y = 4 * y$

$Y = 4 y$

The percentage change is then calculated as:

$\%Change = \frac{Y - y}{y} * 100\%$

$\%Change = \frac{4y - y}{y} * 100\%$

$\%Change = \frac{3y}{y} * 100\%$

$\%Change = 3 * 100\%$

$\%Change = 300\%$

<em>The percentage in y is 300%</em>

7 0

2 years ago

George wishes to increase the percent of acid in 50 ml of a 15% acid solution in water

PtichkaEL [24]

Answer:

screw this question

Step-by-step explanation:

To change from 15% to 25% he needs to add 10%.

If the solution is 50ml then he needs to add 10% which is 5ml... But if he adds 5 ml then the solution will be 55ml total, so that doesn't make sense. He has to take some of the solution that's not acid before adding the 5ml...

3 0

1 year ago

Read 2 more answers